Chapter 9.2, Problem 35E

(a)

To determine

To find:The standard form of the equation of ellipse $x^2+9y^2=36$ .

Answer to Problem 35E

The standard form of the equation of ellipse $x^2+9y^2=36$ is $\frac{x^2}{6^2}+\frac{y^2}{2^2}=1$ .

Explanation of Solution

Given information:

The equation of the ellipse is $x^2+9y^2=36$ .

Calculation:

Rewrite the given equation of ellipse.

  $\begin{array}{c}{x}^2+9y^2=36\\ \frac{x^2}{36}+\frac{9y^2}{36}=1\\ \frac{x^2}{36}+\frac{y^2}{4}=1\\ \frac{x^2}{6^2}+\frac{y^2}{2^2}=1\end{array}$

Therefore, the standard form of the equation of ellipse $x^2+9y^2=36$ is $\frac{x^2}{6^2}+\frac{y^2}{2^2}=1$ .

(b)

To determine

To find:The center, vertices, foci and eccentricity of the ellipse $x^2+9y^2=36$ .

Answer to Problem 35E

The center, vertices, foci and eccentricity of the ellipse $x^2+9y^2=36$ are $\left(0,0\right)$ , $\left(\pm 6,0\right)$ , $\left(\pm 4\sqrt{2},0\right)$ and $\frac{2\sqrt{2}}{3}$ respectively.

Explanation of Solution

Given information:

The equation of the ellipse is $x^2+9y^2=36$ .

Calculation:

As calculated in part(a), the standard form of the equation of ellipse is $\frac{x^2}{6^2}+\frac{y^2}{2^2}=1$ .

Compare the equation of ellipse $\frac{x^2}{6^2}+\frac{y^2}{2^2}=1$ with general equation $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

This gives $a=6$ , $b=2$ , $h=0$ and $k=0$ . The major axis of the ellipse is horizontal.

Use the formula.

  $\begin{array}{c}{c}^2=a^2-b^2\\ c^2=36-4\\ c^2=32\\ c=4\sqrt{2}\end{array}$

Use the formula for eccentricity.

  $\begin{array}{c}e=\frac{c}{a}\\ =\frac{4\sqrt{2}}{6}\\ =\frac{2\sqrt{2}}{3}\end{array}$

The center of the ellipse is $\left(h,k\right)=\left(0,0\right)$ .

The vertices of the ellipse are:

  $\begin{array}{c}\left(h\pm a,k\right)=\left(0\pm 6,0\right)\\ =\left(\pm 6,0\right)\end{array}$

The foci of the ellipse are:

  $\begin{array}{c}\left(h\pm c,k\right)=\left(0\pm 4\sqrt{2},0\right)\\ =\left(\pm 4\sqrt{2},0\right)\end{array}$

The eccentricity of ellipse is $e=\frac{2\sqrt{2}}{3}$ .

Therefore, The center, vertices, foci and eccentricity of the ellipse $x^2+9y^2=36$ are $\left(0,0\right)$ , $\left(\pm 6,0\right)$ , $\left(\pm 4\sqrt{2},0\right)$ and $\frac{2\sqrt{2}}{3}$ respectively.

(c)

To determine

To graph:The ellipse of equation $x^2+9y^2=36$ .

Answer to Problem 35E

The graph of the ellipse $x^2+9y^2=36$ is shown in Figure(1).

Explanation of Solution

Given information:

The equation of the ellipse is $x^2+9y^2=36$ .

Calculation:

The graph of the ellipse $x^2+9y^2=36$ is shown below.

  

Figure(1)

Verify the result graphically:

The graph of the equation $x^2+9y^2=36$ with the help of graphing utility.

  

Figure(2)

As both the graphs gives the same values at each point.

Therefore, both the graphs are similar.